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BigorU [14]
3 years ago
15

Which one is this please help thank u

Mathematics
2 answers:
GrogVix [38]3 years ago
7 0

Answer:

isosceles

Step-by-step explanation:

Two sides are the same length as indicated by the lines on the sides.

This tells use that the triangle is isosceles

postnew [5]3 years ago
6 0

Answer:

Hi, this an isoceles triangle

Step-by-step explanation:

An isoceles triangle has 2 equal sides

A right angled one has a right angle (90 degrees)

An equilateral traingle has all of his sides equal

A scalene has all of his sides different.

So this must be an isoceles triangle because there is basically a proof of the two equal sides (the bars).

Hope its clear!

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What is 0.08% written as a decimal?
otez555 [7]

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0.0008

Step-by-step explanation:

divide 0.08 by 100

answer is 0.0008

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4 years ago
What is the 7-intercept of 2x-4y=12
Bingel [31]

Answer:

-3

Step-by-step explanation:

2x - 4y = 12

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8 0
4 years ago
Read 2 more answers
A lacrosse player throws a ball into the air from a height of 6 feet with an initial vertical velocity of 64 feet per second. Wh
-Dominant- [34]

Answer:

Step-by-step explanation:

I'm going to use calculus to solve this, because it's the simplest way.  

The acceleration due to gravity in feet is the second derivative of the position function.  We will start with the acceleration and work backwards with antiderivatives to get to the position function.

a(t) = -32.  Going backwards and using the fact that the initial vertical velocity is 64 ft/sec, our velocity function is

v(t) = -32t + 64.  Going backwards and using the fact that the initial height of the ball is 6 feet, our position function is

s(t)=-16t^2+64t+6

The first part of this question asks us the maximum height of the ball.  From Physics, we learn that the maximum height of a projectile is reached when the velocity is 0, which happens to be right where the projectile stops for a nanosecond in the air to turn around and come back down.  We set the velocity function equal to 0 and solve for t.

0 = -32t + 64 and

0 = -32(t - 2).  By the Zero Product Property, either -32 = 0 or t - 2 = 0.  It's obvious that -32 does not equal 0, so t - 2 must equal 0.  Solving this for t:

t - 2 = 0 so

t = 2 seconds.  Since the maximum height is reached at a time of 2 seconds, we plug 2 seconds into the position function to get its position at 2 seconds (which is also the max height of the ball).

s(2)=-16(2)^2+64(2)+6 and

s(2) = -64 + 128 + 6 so

s(2) = 70 feet

Now we want to know when the ball will hit the ground.  "When" is a time value, and we know that the height of the ball on the ground is 0, so we sub in a 0 for s(t) and factor the quadratic.

Using the quadratic formula:

t=\frac{-64+/-\sqrt{4096-4(-16)(6)} }{-32} and

t=\frac{-64+/-\sqrt{4480} }{-32} which gives us the 2 solutions

t=\frac{-64+\sqrt{4480} }{-32} and

t=\frac{-64-\sqrt{4480} }{-32}

Plugging into your calculator, the first t = -.0916500 and the second t = 4.091

We all know that time cannot ever be negative, so our t value is 4.09.

Again, from Physics, we know that a projectile reaches it max height at halfway through its travels, so it just goes to follow logically that if it halfway through its travels at 2 seconds, then it will hit the ground at 4 seconds.  And it does!! How awesome is that?!

4 0
4 years ago
<img src="https://tex.z-dn.net/?f=2%20%5Ctimes%202" id="TexFormula1" title="2 \times 2" alt="2 \times 2" align="absmiddle" class
EastWind [94]
4 (i literally had to think about this cause i thought it was too easy )
6 0
4 years ago
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